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Nov 4, 2015 at 17:17 comment added meh Unless I misunderstand your notation $W_0^d = {A \mbox{ s.t. } h^0(A) \geq 1 } = {C_d)} $ and the expected dimension is d . In addition, plugging r=0 into the formula $g -(g-d+r)(r+1) gives the Brill-Noether number of effective line bundles is d, which is to be 'expected'.
Nov 4, 2015 at 14:49 comment added gradstudent @aginensky, Oh then what is the expected dimension?
Nov 4, 2015 at 14:08 comment added meh One statement in the question is wrong $W^0_d$ has expected (and actual) dimension d.
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Nov 2, 2015 at 15:28 answer added Felipe Voloch timeline score: 2
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Oct 2, 2015 at 15:44 comment added roy smith Have you heard of Prym differentials? They are sections of the tensor product of the canonical bundle with a 2 torsion line bundle. This 2 torsion bundle determines a double cover D of the curve C, and the Prym differentials on C correspond to differentials on D that are skew symmetric for the sheet exchange involution of D. Maybe that pattern generalizes. (Oh and there are always only g-1 indept Prym diferentials on a curve of genus g.)
Oct 2, 2015 at 15:41 comment added gradstudent Thanks @abx, if we consider $W^{g-2}_{2g-2}\setminus W^{g-1}_{2g-2}$ this will be preserved by tensor product by a order 2 line bundle. That's why I had this doubt. How do we prove your claim? Can you give me any leads?
Oct 2, 2015 at 6:05 comment added abx Certainly not. Take for $C$ a hyperelliptic curve: then $W^1_2$ has only one point, which is not preserved by tensor product with any line bundle of order 2. In fact, I believe the situation you describe never happens when $\dim W^r_d<g$.
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